How to derive a differential equation of an ellipse I am quite new to differential equations and derivatives. I want to derive an differential form for equation of an ellipse. If i start with an ordinary ellipse equation
\begin{equation}
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
\end{equation}
How do i derive it then to get this form 
$$
-\frac{dx}{dy} = \frac{a^2}{b^2} \frac{y}{x}
$$
I would need an equation and some brief explanation on the procedure. 
 A: The equation $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \tag{1}$$
has two variables: $\{ x, y \}.$ By "derive," it seems that you mean
$ \frac{dx}{dy}.$
Well, differentiating equation $(1)$ w.r.t $y,$ we get:
$$
\frac{d}{dy} \frac{x^2}{a^2} + \frac{d}{dy} \frac{y^2}{b^2} = \frac{d}{dy} 1
\tag{2}$$
First, note that  $\dfrac{d}{dy} 1 = 0,$  $\dfrac{d}{dy} y = 1,$ and $\dfrac{d}{dy} f^2 = 2 f \dfrac{df}{dy}.$
So


*

*$ \dfrac{d}{dy} \dfrac{x^2}{a^2} = 2 \dfrac{x^{2-1}}{a^2} \dfrac{dx}{dy}$

*$ \dfrac{d}{dy} \dfrac{y^2}{b^2} = 2 \dfrac{y^{2-1}}{b^2} \dfrac{dy}{dy} $
In other words, equation $(2)$ becomes:
$$
 2\frac{x}{a^2} \frac{dx}{dy} + 2 \frac{y}{b^2} = 0.
$$
The rest is simple algebra, you can isolate $\dfrac{dx}{dy}$ one side, and get:
$$ -\frac{dx}{dy} = \frac{a^2}{b^2} \frac{y}{x} $$
A: The answer you want is actually not the differential equation of the family of ellipse. A differential equation is free of arbitrary constants like $a$ and $b$. Since there are two arbitrary constants, you need to differentiate 2 times (the order of the differential equation should be 2). The answer you want is just the negative of slope of normal at any point $(x,y)$ on the ellipse.
